A manufacturer can sell x items per day at a price p rupees each, where `p=125-(5)/(3)x`. The cost of production for x items is `500 + 13x + 0.2x^(2)`. Find how much he should produce have a maximum profit, assuming all items produces are sold.

Answer

To find the number of items \( x \) that the manufacturer should produce to achieve maximum profit, we can follow these steps:

### Step 1: Determine the Selling Price
The selling price \( p \) of each item is given by the equation:
\[ p = 125 - \frac{5}{3}x \]
Thus, the total selling price \( S \) for \( x \) items is:
\[ S = x \cdot p = x \left(125 - \frac{5}{3}x\right) = 125x - \frac{5}{3}x^2 \]

### Step 2: Determine the Cost Price
The cost of production \( C \) for \( x \) items is given by:
\[ C = 500 + 13x + 0.2x^2 \]

### Step 3: Determine the Profit Function
The profit \( P \) is calculated as the difference between the total selling price and the total cost price:
\[ P = S - C = \left(125x - \frac{5}{3}x^2\right) - \left(500 + 13x + 0.2x^2\right) \]
Simplifying this gives:
\[ P = 125x - \frac{5}{3}x^2 - 500 - 13x - 0.2x^2 \]
Combining like terms:
\[ P = (125 - 13)x - \left(\frac{5}{3} + 0.2\right)x^2 - 500 \]
\[ P = 112x - \left(\frac{5}{3} + \frac{2}{10}\right)x^2 - 500 \]
To combine the coefficients of \( x^2 \), we convert \( 0.2 \) to a fraction:
\[ 0.2 = \frac{2}{10} = \frac{1}{5} \]
Finding a common denominator (15):
\[ \frac{5}{3} = \frac{25}{15}, \quad \frac{1}{5} = \frac{3}{15} \]
Thus,
\[ \frac{5}{3} + 0.2 = \frac{25}{15} + \frac{3}{15} = \frac{28}{15} \]
So the profit function becomes:
\[ P = 112x - \frac{28}{15}x^2 - 500 \]

### Step 4: Differentiate the Profit Function
To find the maximum profit, we differentiate \( P \) with respect to \( x \):
\[ \frac{dP}{dx} = 112 - \frac{56}{15}x \]

### Step 5: Set the Derivative to Zero
Setting the derivative equal to zero to find critical points:
\[ 112 - \frac{56}{15}x = 0 \]
Solving for \( x \):
\[ \frac{56}{15}x = 112 \]
\[ x = 112 \cdot \frac{15}{56} = 30 \]

### Step 6: Verify Maximum Profit
To confirm that this is a maximum, we can check the second derivative:
\[ \frac{d^2P}{dx^2} = -\frac{56}{15} \]
Since this is negative, it indicates that the profit function has a maximum at \( x = 30 \).

### Conclusion
The manufacturer should produce **30 items** to achieve maximum profit.

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